I would advocated that readability must be a primer for us as developers.

Apache Commons Lang (commons-lang) provides a simple fluent utility which reads a lot clearer:

if (is(obj1).greaterThan(obj2)) {
     // do something
}

Note: is is shorthand for ComparableUtils.is which can be imported the following this static import statement:

import static org.apache.commons.lang3.compare.ComparableUtils.is;

Bounded type parameters are key to the implementation of generic algorithms. Consider the following method that counts the number of elements in an array T[] that are greater than a specified element elem.

public static <T> int countGreaterThan(T[] anArray, T elem) {
    int count = 0;
    for (T e : anArray)
        if (e > elem)  // compiler error
            ++count;
    return count;
}

The implementation of the method is straightforward, but it does not compile because the greater than operator (>) applies only to primitive types such as short, int, double, long, float, byte, and char. You cannot use the > operator to compare objects. To fix the problem, use a type parameter bounded by the Comparable<T> interface:

public interface Comparable<T> {
    public int compareTo(T o);
}

The resulting code will be:

public static <T extends Comparable<T>> int countGreaterThan(T[] anArray, T elem) {
    int count = 0;
    for (T e : anArray)
        if (e.compareTo(elem) > 0)
            ++count;
    return count;
}

bounded type parameters

The compareTo method is described as follows:

Compares this object with the specified object for order. Returns a negative integer, zero, or a positive integer as this object is less than, equal to, or greater than the specified object.

Let's say we would like to compare Jedis by their age:

class Jedi implements Comparable<Jedi> {

    private final String name;
    private final int age;
        //...
}

Then if our Jedi is older than the provided one, you must return a positive, if they are the same age, you return 0, and if our Jedi is younger you return a negative.

public int compareTo(Jedi jedi){
    return this.age > jedi.age ? 1 : this.age < jedi.age ? -1 : 0;
}

By implementing the compareTo method (coming from the Comparable interface) your are defining what is called a natural order. All sorting methods in JDK will use this ordering by default.

There are ocassions in which you may want to base your comparision in other objects, and not on a primitive type. For instance, copare Jedis based on their names. In this case, if the objects being compared already implement Comparable then you can do the comparison using its compareTo method.

public int compareTo(Jedi jedi){
    return this.name.compareTo(jedi.getName());
}

It would be simpler in this case.

Now, if you inted to use both name and age as the comparison criteria then you have to decide your oder of comparison, what has precedence. For instance, if two Jedis are named the same, then you can use their age to decide which goes first and which goes second.

public int compareTo(Jedi jedi){
    int result = this.name.compareTo(jedi.getName());
    if(result == 0){
        result = this.age > jedi.age ? 1 : this.age < jedi.age ? -1 : 0;
    }
    return result;
}

If you had an array of Jedis

Jedi[] jediAcademy = {new Jedi("Obiwan",80), new Jedi("Anakin", 30), ..}

All you have to do is to ask to the class java.util.Arrays to use its sort method.

Arrays.sort(jediAcademy);

This Arrays.sort method will use your compareTo method to sort the objects one by one.